geometryinfinitysoft-question
I know the line segment have a infinite number of points, but i know that exist different kinds of infinity ( $\aleph_0 $). My question is there same number of points on segment of line and entire line.
If you know some good book or article about this topic i will be great full.
Sorry for bad English.
The equation of a line through $(x_A,y_A)$ and $(x_B,y_B)$ is
$$A \mathbf{x} + B \mathbf{y} + C=0\\(y_B-y_A) \mathbf{x} + (x_A-x_B) \mathbf{y} + x_B y_A - x_A y_B = 0 $$
The equation of parallel passing through $(x_C,y_C)$ is
$$A \mathbf{x} + B \mathbf{y} + C'=0 \\(y_B-y_A) \mathbf{x} + (x_A-x_B) \mathbf{y} + y_C (x_B-x_A) + x_C (y_A-y_B) = 0 $$
The point on the parallel a line $A \mathbf{x} + B \mathbf{y} + C' = 0$ closest to point $(x_A,y_A)$ is
$$ (x_A',y_A') = \left( \frac{B^2 x_A-A (B y_A+C')}{A^2+B^2},\; \frac{A^2 y_A-B (A x_A+C')}{A^2+B^2} \right) $$
and similarly for the other point
$$ (x_B',y_B') = \left( \frac{B^2 x_B-A (B y_B+C')}{A^2+B^2},\; \frac{A^2 y_B-B (A x_B+C')}{A^2+B^2} \right) $$
with
$$\begin{align} A & = y_B-y_A = -13\\ B & = x_A-x_B = -25 \\ C & = x_B y_A - x_A y_B = 630 \\ C' & = y_C (x_B-x_A) + x_C (y_A-y_B) = 406 \end{align} $$
$$ (x_A',y_A') = \left( \frac{2514}{397}, \; \frac{5140}{397} \right) = (6.33, 12.95) $$ $$ (x_B',y_B') = \left( \frac{12439}{397}, \; -\frac{21}{397} \right) = (31.33, -0.05)$$
Confirmation with Geogebra
You seem to be on the right track for the first part. Let me see if I can put it into more precise terms: that of lines in a geometric plane.
Take some line $L$ in the plane, and pick a point $\langle x_0,y_0\rangle$ of $L$ to be the "origin." Now, there are two points of $L$ that are a distance of $1$ away from $\langle x_0,y_0\rangle$--that is, there are two $\langle x,y\rangle\in L$ such that $$\sqrt{(x-x_0)^2+(y-y_0)^2}=1.$$ Pick one of them to give the direction of our "positive $x$-axis," say $\langle x_1,y_1\rangle.$ Letting $$\langle x^+,y^+\rangle=\langle x_1-x_0,y_1-y_0\rangle,$$ we then have $$\sqrt{(x^+)^2+(y^+)^2}=1,$$ and the points of $L$ are exactly the points of the form $$\langle x_t,y_t\rangle:=\langle x_0,y_0\rangle+t\langle x^+,y^+\rangle=\langle x_0+tx^+,y_0+ty^+\rangle$$ for some real number $t.$ (Note that this notation is unambiguous when we let $t=0$ or $t=1.$)
At this point, you may be wondering why I did this. Time for the slick trick. Given two points $\langle x_s,y_s\rangle$ and $\langle x_t,y_t\rangle,$ we define $\langle x_s,y_s\rangle\oplus\langle x_t,y_t\rangle:=\langle x_{s+t},y_{s+t}\rangle$ and $\langle x_s,y_s\rangle\odot\langle x_t,y_t\rangle:=\langle x_{s\cdot t},y_{s\cdot t}\rangle.$ In this fashion, we give the line $L$ an arithmetic structure just like the one that the real numbers have. Next, we say that $\langle x_s,y_s\rangle\lessdot\langle x_t,y_t\rangle$ if and only if $s<t,$ thereby giving $L$ an order structure just like the one that the reals have. Now, note that $$\begin{eqnarray}\sqrt{(x_s-x_t)^2+(y_s-y_t)^2} & = & \sqrt{(sx^+-tx^+)^2+(sy^+-ty^+)^2}\\ & = & \sqrt{\bigl((s-t)x^+\bigr)^2+\bigl((s-t)y^+\bigr)^2}\\ & = & \sqrt{(s-t)^2(x^+)^2+(s-t)^2(x^+)^2}\\ & = & \sqrt{(s-t)^2\left((x^+)^2+(y^+)^2\right)}\\ & = & \sqrt{(s-t)^2}\sqrt{(x^+)^2+(y^+)^2}\\ & = & \sqrt{(s-t)^2}\\ & = & |s-t|.\end{eqnarray}$$ Putting this into words, the distance between the points $\langle x_s,y_s\rangle$ and $\langle x_t,y_t\rangle$ is exactly the distance between the real numbers $s$ and $t.$ In this fashion, we have shown that $L$ is a "natural copy" of the real number line! So, when considering a statement about a line in the plane, we may as well consider the equivalent statement about the real number line, for simplicity.
Now, let me start addressing the issues I see.
[B]y just choosing a point you can distinguish between "1" or "2" infinities. Both infinite, yet choosing a point "divides" (I know I can't do that) the original one in seemingly 2 different infinites. Which actually "both" have the same properties as the one we started with.
This is partly true! By choosing our "origin", we partitioned ("divided") $L$ into three separate pieces: (i) the "origin," (ii) the "positive $x$-axis," and (iii) the "negative $x$-axis." Translated into terms of real numbers, every real number is exactly one of the following: (i) $0,$ (ii) a positive number, or (iii) a negative number. And, indeed, the set of positive numbers and the set of negative numbers have the same cardinality (roughly, "number of points") as the real number line, and also the same measure (roughly, "length").
However, there is a key property that these two do not share with the real numbers. For example, given any real number $s,$ we can find a real number $t$ less than $s$ such that $|s-t|$ is as large as we like. We cannot do this with the positive numbers. Letting $s$ be some positive number, we have for any positive number $t$ less than $s$ that $|s-t|<|s|.$ Put into more intuitive terms, if we pick a positive point, then the positive numbers extend infinitely from it in one direction, but finitely in the other! Similarly, the negative numbers only extend infinitely in one direction from any given negative point.
So if I choose 0 points on that line, that line is composed of an infinite amount of points.
- 1 point to get 2 infinite lines.
- 2 points and we have 3 infinite lines
- 3 points and we have 4..
- 4 points and we have 5..
Again, this is partly true. If $n$ is a nonnegative integer, and we choose $n$ distinct points on the line, then we partition the line into $n$ points and $n+1$ infinite sets, and it can be shown that each of the infinite sets has the same cardinality as the real number line. However, the term "lines" is a misnomer, here!
In the case $n=1,$ we saw in the discussion above that we have two infinite rays. More generally, whenever $n\ge1,$ the partition will have $n$ points, $2$ infinite rays, and $n-1$ open line segments. Open segments extend infinitely in neither direction! Furthermore, unlike rays and lines, open segments have finite measure!
Now, if we look at a given open segment $I$ of the real numbers, we can define an increasing function $f$ on $I$ that matches up all the points of $I$ in a one-to-one fashion with all the points of the real number line, then use the function $f$ to define a brand new "distance" function on $I$ that makes $I$ of infinite "length." However, this is not the same as the usual distance function. Also, the definition of this "distance" function depends entirely on our choice of the function $f,$ and we can choose $f$ in infinitely-many ways! What this means is that this notion of "distance" isn't "natural," since there is no canonical way to choose $f$. (We can similarly create an artificial distance function for any open ray $R$ of the real line, which will allow $R$ to "extend infinitely" in both directions, as far as that artificial distance function is concerned.)
Now, if we discard considerations of length, and stick to cardinality alone, then everything else you said prior to this next paragraph is true.
So now let's say I take an infinite amount of points, I would still have an infinite amount of lines of infinite length between an infinite amount of points. All with the same properties as the first infinity we started with.
At this point, everything falls apart. It can work (if we change "lines of infinite length" to "infinite sets"), but it may not. If the points we pick out are the integers, say, then we have an infinite number of open segments (of length $1$) in our partition, and it is true that they each have the same cardinality as the real numbers. However, if the points we pick are the rational numbers, for example, then we no longer have any open intervals (that is: open rays, open segments, or lines) in our partition, because the rational numbers are dense in the reals (meaning that every open interval of the real line contains a rational number), so our partition contains only points!
But (you may ask) how can this be? After all, if I remove $n$ points from the line, then I have $n+1$ open intervals that have the same cardinality as the line! So, by induction, shouldn't I be able to remove infinitely-many points and have infinitely-many open intervals left over? Well, no. What induction allows us to show is that it works for every finite number of points that we remove. This is just one of many examples that demonstrate that induction doesn't let us extend the result "to infinity" (and beyond).
Let me know if there's anything you didn't understand about this answer, or if you just want to verify that you've understood it correctly.
Best Answer
If you mean the cardinality of $L$, when you say "number" of points on $L$, then the number of points on a segment of a line and the entire line is same.
For example, consider $f:(-1,1) \to \mathbb{R}$ defined as $f(x) = \tan {\frac{\pi x}{2}}$. $f(x)$ is both one-one and onto.